Let $A$ be $m \times n$ matrix such that $A = UDV^T$ where $V$ is orthogonal $n \times n$ matrix that contains the eigenvectors of $A^TA$ and $U$ is orthogonal $m \times m$ matrix. Let $A$ have $k \le n$ positive singular values. Then $\{v_1, \ldots, v_k\}$ form an orthonormal basis for the row space of $A$.
Since $U, V$ are orthogonal, they are invertible. For example, $U^T = U^{-1}$ meaning the inverse for $U$ exists. Then rank$(A) =$ rank$(UDV^T) =$ rank$(DV^T) = $rank$(D).$ Since $D$ has $k$ linearly independent row vectors, rank$(A) =$ rank$(D) = k.$ Since rank$(A) = $ row rank$(A)$, we have that the dimension of the row space is $k.$
I am having difficulty showing $k$ $v_i$'s are in the row space. How do we go about that?